Answer
The value of $\underset{x\to 0}{\mathop{\lim }}\,\frac{\sqrt{{{x}^{2}}+9}-3}{{{x}^{2}}}$ is $\frac{1}{6}$.
Work Step by Step
Consider the function $ f\left( x \right)=\frac{\sqrt{{{x}^{2}}+9}-3}{{{x}^{2}}}$.
Simplify the above function,
Multiplying the numerator and the denominator of the function by $\sqrt{{{x}^{2}}+9}+3$,
So,
$\begin{align}
& f\left( x \right)=\frac{\sqrt{{{x}^{2}}+9}-3}{{{x}^{2}}}\left( \frac{\sqrt{{{x}^{2}}+9}+3}{\sqrt{{{x}^{2}}+9}+3} \right) \\
& =\frac{{{x}^{2}}+9-9}{{{x}^{2}}\left( \sqrt{{{x}^{2}}+9}+3 \right)} \\
& =\frac{{{x}^{2}}}{{{x}^{2}}\left( \sqrt{{{x}^{2}}+9}+3 \right)} \\
& =\frac{1}{\sqrt{{{x}^{2}}+9}+3}
\end{align}$
The function $ g\left( x \right)={{x}^{2}}+9$ is a polynomial.
Now, find the value of $\underset{x\to 0}{\mathop{\lim }}\,\frac{\sqrt{{{x}^{2}}+9}-3}{{{x}^{2}}}$,
$\begin{align}
& \underset{x\to 0}{\mathop{\lim }}\,\frac{\sqrt{{{x}^{2}}+9}-3}{{{x}^{2}}}=\underset{x\to 0}{\mathop{\lim }}\,\frac{1}{\sqrt{{{x}^{2}}+9}+3} \\
& =\frac{1}{\sqrt{\underset{x\to 0}{\mathop{\lim }}\,{{x}^{2}}+9}+3} \\
& =\frac{1}{\sqrt{0+9}+3} \\
& =\frac{1}{\sqrt{9}+3}
\end{align}$
$\begin{align}
& =\frac{1}{3+3} \\
& =\frac{1}{6} \\
\end{align}$
Thus, the value of $\underset{x\to 0}{\mathop{\lim }}\,\frac{\sqrt{{{x}^{2}}+9}-3}{{{x}^{2}}}$ is $\frac{1}{6}$.
